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<h3 class="heading"><span class="type">Paragraph</span></h3>
<p>Then <span class="process-math">\((\ref{eq5_22})_2\)</span> becomes</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\sum_{n=0}^{\infty}[n (n-1)d_n+d_{n-2}] x^n=0,
\end{equation*}
</div>
<p class="continuation">which gives the recurrence relation</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\begin{aligned}
&amp;d_n=-\frac{d_{n-2}}{n(n-1)},\quad n\geq 2\\
&amp; d_{2k}=(-1)^k \frac{1}{(2 k)!},\quad d_{2k+1}=(-1)^k \frac{d_1}{(2k+1)!},
\end{aligned}
\end{equation*}
</div>
<p class="continuation">where <span class="process-math">\(d_1\)</span> is arbitrary. We set <span class="process-math">\(d_1=0\)</span> so that <span class="process-math">\(d_{2k+1}=0\text{.}\)</span> Then</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\begin{aligned}
y_2 &amp;=a y_1 \ln x+\sum_{n=0}^{\infty} d_n x^{n-1/2}\\
&amp;=0+x^{-1/2} \left[ \sum_{k=0}^{\infty} d_{2k} x^{2k}+\sum_{k=0}^{\infty} d_{2k+1} x^{2k+1}   \right]\\
&amp;=x^{-1/2} \sum_{k=0}^{\infty} (-1)^k \frac{x^{2 k}}{(2 k)!}=x^{-1/2} \cos x.
\end{aligned}
\end{equation*}
</div>
<span class="incontext"><a href="sec5_5.html#p-240" class="internal">in-context</a></span>
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